Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 8 - Further Applications of Integration - 8.1 Arc Length - 8.1 Exercises: 10

Answer

$L = \int^{3}_{2} \sqrt{[\frac{1}{2}(x^{2} -2)]^2}dx = \frac{13}{6}$

Work Step by Step

$36y^{2} = (x^2 -4)^{3}, y\geq0$ then $y = 1/6 (x^{2} -4)^{3/2} $ and $dy/dx = \frac{1}{6} \frac{3}{2} (x^{2} -4)^{1/2}(2x) = \frac{1}{2}x(x^{2}-4)^{1/2}$ $1+(dy/dx)^{2} = 1+(1/4)x^2(x^2-4) = (1/4) x^{4} - x^{2} +1 = (1/4)(x^{4} -4x^{2} +4) = [(1/2) (x^{2} - 2)]^2$ Therefore $L = \int^{3}_{2} \sqrt{[\frac{1}{2}(x^{2} -2)]^2}dx = \frac{1}{2}[(9-6)-(\frac{8}{3}-4)] = \frac{13}{6}$
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